Question

In the figure given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.

Given that AB = 8 cm, calculate PQ.

• A. 16 cm
• B. 15 cm
• C. 19 cm
• D. 25 cm

Answer 1

The correct option is B

15 cm

In the figure, two circles with centres P and Q and radii 6 cm and 3 cm respectively

ABC is the common transverse tangent to the two circles. AB = 8 cm

Join AP and CQ

$\because$ AC is the tangent to the two circles and PA and QC are the radii

By Theorem- The tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore PA\perp AC$ and $QC\perp AC$

Now in right $\Delta APB,$
Applying pythagoras theorem,

$(PB{)}^2=A{P}^2+A{B}^2=(6{)}^2+(8{)}^2$

= 36+64= 100= $(10{)}^2$

$\therefore$ PB = 10 cm

Similarly in $\Delta APB$ and $\Delta CBQ,$

$\angle A=\angle C$ (each ${90}^0)$

$\angle ABP=\angle CBQ$

(vertically opposite angles)

$\therefore \Delta ABP\sim \Delta CBQ$ (AA axiom)

$\therefore \frac{AB}{CB}=\frac{AP}{CQ}=\frac{PB}{BQ}$

$\Rightarrow \frac{8}{BC}=\frac{6}{3}\Rightarrow BC=\frac{8\times 3}{6}$ = 4cm

and $\frac{10}{BQ}=\frac{6}{3}\Rightarrow BQ=\frac{10\times 3}{6}$= 5 cm

$\therefore$ PQ = PB+BQ = 10+5 = 15 cm